Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.) 2 sin2(θ) = 2 + cos(2θ)
θ=

Use a Double- or Half-Angle Formula to solve the equation in the interval [0, 2π). (Enter your answers as a comma-separated list.)
cos(2θ) + 7 cos(θ) = 8
θ=

For the first equation, recall that sin²(θ) = (1 - cos(2θ))/2. Then

2 sin²(θ) = 2 + cos(2θ)

1 - cos(2θ) = 2 + cos(2θ)

2 cos(2θ) = -1

cos(2θ) = -1/2

2θ = arccos(-1/2) + 2nπ   or   2θ = 2π - arccos(-1/2) + 2nπ

(where n is any integer)

2θ = 2π/3 + 2nπ   or   2θ = 4π/3 + 2nπ

θ = π/3 + nπ   or   θ = 2π/3 + nπ

In the interval [0, 2π), the solutions are θ = π/3, 2π/3, 4π/3, 5π/3.

For the second equation, rearrange the previous identity to arrive at

cos(2θ) = 1 - 2 sin²(θ) = 2 cos²(θ) - 1

Then

cos(2θ) + 7 cos(θ) = 8

2 cos²(θ) - 1 + 7 cos(θ) = 8

2 cos²(θ) + 7 cos(θ) - 9 = 0

(2 cos(θ) + 9) (cos(θ) - 1) = 0

2 cos(θ) + 9 = 0   or   cos(θ) - 1 = 0

cos(θ) = -9/2   or   cos(θ) = 1

Since |-9/2| > 1, and cos(θ) is bounded between -1 and 1, the first case offers no solutions. This leaves us with

cos(θ) = 1

θ = arccos(1) + 2nπ

θ = 2nπ

so that there is only one solution in [0, 2π), θ = 0.

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step-by-step explanation:

not enough information

step-by-step explanation:

we cannot know this, because the other students could have handed out 10 flyers, or they could have handed out 100. one way, the mean would go up, and another way, the mean would go down.